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Let $(x_{n})$ be defined recursively by $x_{n+1}=\sqrt{a+x_{n}}$ where $a>0$ and $x_1=\sqrt{a}$. I've shown that $(x_{n})$ is bounded above by $1+\sqrt{a}$. If I can show it's increasing, then it converges by the Monotone Convergence Theorem, and I have to find the limit. I'm having trouble showing it's increasing though: $x_n \leq x_{n+1}$ iff $x_n \leq \sqrt{a+x_n}$. We can show inductively that $0 \leq x_n$ for all natural numbers $n$, so I can square both sides and get $(x_n)$ is increasing iff $x_n^2-x_n-a \leq 0$ for all $n$, which isn't true. What am I doing wrong? I say that it isn't true because $x_n^2$ is the dominant term, so I can pick an N large enough s.t. $x_N^2-x_N-a>0$

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    $\begingroup$ why isn't it true? It holds for $x_1$. Is there a problem after that? I haven't done any other calculations though. $\endgroup$ – voldemort Jan 30 '14 at 7:27
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    $\begingroup$ $x_n^2$ is the dominant term, so if I take N large enough, then $x_N^2-x_N-a > 0$ $\endgroup$ – user124910 Jan 30 '14 at 7:30
  • $\begingroup$ Ah, never mind... I realized what was confusing me. I was thinking of this as a quadratic in $n$ and not $x_n$ lol. $\endgroup$ – user124910 Jan 30 '14 at 7:36
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Assume you know $x_{n}<x_{n-1}$ (which you know for $n=1$ if one defines $x_0=0$ (whch fits the recursion). Then conclude $x_{n+1}\sqrt{a+x_n}>\sqrt{a+x_{n-1}}=x_n$.

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